Question

Let \(f(x, y)=x\) and let \(C\) be the segment of the parabola \(y=x^{2}\) joining \(O(0,0)\) and \(P(1,1)\) a. Find a parameterization of \(C\) in the direction from \(O\) to \(P\). Evaluate \(\int_{\mathcal{C}} f d s\) b. Find a parameterization of \(C\) in the direction from \(P\) to \(O\) Evaluate \(\int_{\mathcal{C}} f d s\) c. Compare the results of parts (a) and (b).

Short answer

In summary, when evaluating the line integral of the function \(f(x, y) = x\) along the curve \(C\) (a segment of parabola \(y = x^2\)) both in the direction from point \(O(0, 0)\) to point \(P(1, 1)\) and in the opposite direction from point \(P\) to point \(O\), we find that their magnitudes are the same but they have opposite signs. This implies that the orientation of the curve matters in evaluating line integrals for this function.

Step-by-step solution

Parameterize C in the direction from O to P

We can parameterize the curve by introducing a parameter \(t\) which ranges from 0 (at point \(O\)) to 1 (at point \(P\)). The parameterization of the curve \(C\) can be given by the vector-valued function \(\vec{r}(t)=\begin{bmatrix}t\\t^2\end{bmatrix}\) for \(0\leq t \leq 1.\) b. From \(P\) to \(O\):

Parameterize C in the direction from P to O

Now, we need to reverse the direction from \(P\) to \(O\). We can do this by letting the parameter move from 1 to 0. In this case, the new parameter is \(1 - t\), and the new parameterization of the curve \(C\) can be given by the vector-valued function \(\vec{r}(t)=\begin{bmatrix}1-t\\(1-t)^2\end{bmatrix}\) for \(0\leq t\leq 1.\) ##Step 2: Compute the magnitude of the derivative of the position vector## a. From \(O\) to \(P\):

Compute the derivative and its magnitude from O to P

First, we compute the derivative \(\frac{d\vec{r}}{dt}\): \(\frac{d\vec{r}}{dt} = \begin{bmatrix}1\\2t\end{bmatrix}\) Next, we compute the magnitude of the derivative: \(|\frac{d\vec{r}}{dt}| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2}\) b. From \(P\) to \(O\):

Compute the derivative and its magnitude from P to O

First, we compute the derivative \(\frac{d\vec{r}}{dt}\): \(\frac{d\vec{r}}{dt} = \begin{bmatrix}-1\\-2(1-t)\end{bmatrix}\) Next, we compute the magnitude of the derivative: \(|\frac{d\vec{r}}{dt}| = \sqrt{(-1)^2 + (-2(1-t))^2} = \sqrt{1 + 4(1-t)^2}\) ##Step 3: Evaluate the line integral## a. From \(O\) to \(P\):

Line integral from O to P

Now, we can evaluate the line integral: \(\int_{\mathcal{C}} f ds = \int_{0}^{1} f(\vec{r}(t)) |\frac{d\vec{r}}{dt}| dt\) \(\int_{\mathcal{C}} f ds = \int_{0}^{1} (t) \sqrt{1+4t^2} dt\) To evaluate this integral, let's use substitution method: \(u = 1 + 4t^2\), then \(\frac{du}{dt} = 8t\) or \(\frac{1}{8}du = t dt\). The new limits of integration are \(u = 1\) (for \(t = 0\)) and \(u=5\) (for \(t=1\)): \(\int_{\mathcal{C}} f ds = \int_{1}^{5} \frac{1}{8}\sqrt{u} du\) Now the integral is simpler and can be solved easily: \(\int_{\mathcal{C}} f ds = \frac{1}{12}(u^{3/2})\Big|_{1}^{5} = \frac{1}{12}(5\sqrt{5}-1)\) b. From \(P\) to \(O\):

Line integral from P to O

Now we evaluate the line integral for the reverse direction: \(\int_{\mathcal{C}} f ds = \int_{0}^{1} f(\vec{r}(t)) |\frac{d\vec{r}}{dt}| dt\) \(\int_{\mathcal{C}} f ds = \int_{0}^{1} (1-t) \sqrt{1+4(1-t)^2} dt\) You can use substitution method again and will find: \(\int_{\mathcal{C}} f ds = \frac{1}{12}(u^{3/2})\Big|_{5}^{1} = \frac{1}{12}(1-5\sqrt{5})\) c. Compare the results:

Compare the results of parts (a) and (b)

From part (a) and part (b), we found that the line integral from \(O\) to \(P\) is equal to: \(\frac{1}{12}(5\sqrt{5}-1)\) while the line integral from \(P\) to \(O\) is equal to: \(\frac{1}{12}(1-5\sqrt{5})\) Their magnitudes are the same, but they have opposite signs. This means that the integration evaluated in opposite directions along the curve \(C\) gives opposite values.

Key concepts

Parameterization

When dealing with line integrals, parameterization is a key step. To parameterize a curve, you introduce a variable, such as the parameter \( t \), to describe each point on the curve in terms of functions of \( t \). This approach is essential for translating the curve into a form that can be integrated. For a curve \( C \), representing a segment of the parabola \( y = x^2 \), we can express the curve using the parameter \( t \) to define our coordinate points.
For instance, you can use the vector-valued function \( \vec{r}(t) = \begin{bmatrix} t \ t^2 \end{bmatrix} \) when moving from the point \( O(0,0) \) to \( P(1,1) \).
This form captures each point on the curve as \( t \) varies from 0 to 1.
Each point is expressed as \( x = t \) and \( y = t^2 \), matching the parabola's equation \( y = x^2 \). This parameterization sets the stage for further analysis and integration.

Vector-Valued Functions

A vector-valued function is a function that takes a parameter or a set of parameters and returns a vector. These functions are crucial for describing curves in higher dimensions, like the two-dimensional plane or three-dimensional space.
In the example of parameterizing the parabola from point \( O \) to point \( P \), the function \( \vec{r}(t) = \begin{bmatrix} t \ t^2 \end{bmatrix} \) is a vector-valued function.
It maps each value of \( t \) to a corresponding point on the curve, expressed as a vector.
The function allows us to work with the curve geometrically, providing a framework for calculating quantities like the line integral, which sums a scalar field over the curve.
Vector-valued functions effectively bring the abstract idea of a curve into a calculable format.

Substitution Method

The substitution method is a powerful tool used to simplify integration, especially in line integrals where complex expressions are common.
In line integrals, it often involves changing variables to transform the integral into a form that is easier to evaluate.
For example, when evaluating \( \int_{0}^{1} t \sqrt{1+4t^2} \, dt \), you can set \( u = 1 + 4t^2 \), turning the integral into \( \int_{1}^{5} \frac{1}{8} \sqrt{u} \, du \).
This transformation simplifies both the function inside the integral and the process of integration itself.
The substitution method facilitates easier computation by reducing complex integrals to more manageable terms, making it an indispensable technique in calculus.

Derivatives

Derivatives play a central role in evaluating line integrals, particularly when calculating the magnitude of a curve's position vector as it changes.
To compute the derivative of a vector-valued function like \( \vec{r}(t) = \begin{bmatrix} t \ t^2 \end{bmatrix} \), we differentiate each component with respect to \( t \).
This yields \( \frac{d\vec{r}}{dt} = \begin{bmatrix} 1 \ 2t \end{bmatrix} \) from \( O \) to \( P \).
The magnitude \( |\frac{d\vec{r}}{dt}| = \sqrt{1 + 4t^2} \) represents how fast the curve's position vector is changing.
This magnitude is integral to finding the value of the line integral, as it affects the scaling of the function being integrated. Understanding derivatives is crucial for subsequent calculations, providing insights into the behavior and rate of change of curves in space.